Multiply $ \color{blue}{p} $ by $ \left( p-1\right) $ $$ \color{blue}{p} \cdot \left( p-1\right) = p^2-p $$Multiply $ \color{blue}{p} $ by $ \left( p-1\right) $ $$ \color{blue}{p} \cdot \left( p-1\right) = p^2-p $$

Multiply each term of $ \left( \color{blue}{p^2-p}\right) $ by each term in $ \left( p-2\right) $.

$$ \left( \color{blue}{p^2-p}\right) \cdot \left( p-2\right) = p^3-2p^2-p^2+2p $$

Combine like terms:

$$ p^3 \color{blue}{-2p^2} \color{blue}{-p^2} +2p = p^3 \color{blue}{-3p^2} +2p $$

Multiply $p$ by $ \dfrac{p-1}{2} $ to get $ \dfrac{ p^2-p }{ 2 } $.

Step 1: Write $ p $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} p \cdot \frac{p-1}{2} & \xlongequal{\text{Step 1}} \frac{p}{\color{red}{1}} \cdot \frac{p-1}{2} \xlongequal{\text{Step 2}} \frac{ p \cdot \left( p-1 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ p^2-p }{ 2 } \end{aligned} $$

Multiply $p^2-p$ by $ \dfrac{p-2}{6} $ to get $ \dfrac{p^3-3p^2+2p}{6} $.

Step 1: Write $ p^2-p $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} p^2-p \cdot \frac{p-2}{6} & \xlongequal{\text{Step 1}} \frac{p^2-p}{\color{red}{1}} \cdot \frac{p-2}{6} \xlongequal{\text{Step 2}} \frac{ \left( p^2-p \right) \cdot \left( p-2 \right) }{ 1 \cdot 6 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ p^3-2p^2-p^2+2p }{ 6 } = \frac{p^3-3p^2+2p}{6} \end{aligned} $$

Multiply $p^3-3p^2+2p$ by $ \dfrac{p-3}{24} $ to get $ \dfrac{p^4-6p^3+11p^2-6p}{24} $.

Step 1: Write $ p^3-3p^2+2p $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} p^3-3p^2+2p \cdot \frac{p-3}{24} & \xlongequal{\text{Step 1}} \frac{p^3-3p^2+2p}{\color{red}{1}} \cdot \frac{p-3}{24} \xlongequal{\text{Step 2}} \frac{ \left( p^3-3p^2+2p \right) \cdot \left( p-3 \right) }{ 1 \cdot 24 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ p^4-3p^3-3p^3+9p^2+2p^2-6p }{ 24 } = \frac{p^4-6p^3+11p^2-6p}{24} \end{aligned} $$

Multiply $ \dfrac{p^2-p}{2} $ by $ 4 $ to get $ \dfrac{ 4p^2-4p }{ 2 } $.

Step 1: Write $ 4 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{p^2-p}{2} \cdot 4 & \xlongequal{\text{Step 1}} \frac{p^2-p}{2} \cdot \frac{4}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( p^2-p \right) \cdot 4 }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4p^2-4p }{ 2 } \end{aligned} $$

Multiply $ \dfrac{p^3-3p^2+2p}{6} $ by $ -8 $ to get $ \dfrac{ -8p^3+24p^2-16p }{ 6 } $.

Step 1: Write $ -8 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{p^3-3p^2+2p}{6} \cdot -8 & \xlongequal{\text{Step 1}} \frac{p^3-3p^2+2p}{6} \cdot \frac{-8}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( p^3-3p^2+2p \right) \cdot \left( -8 \right) }{ 6 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -8p^3+24p^2-16p }{ 6 } \end{aligned} $$

Multiply $ \dfrac{p^4-6p^3+11p^2-6p}{24} $ by $ 16 $ to get $ \dfrac{ 16p^4-96p^3+176p^2-96p }{ 24 } $.

Step 1: Write $ 16 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{p^4-6p^3+11p^2-6p}{24} \cdot 16 & \xlongequal{\text{Step 1}} \frac{p^4-6p^3+11p^2-6p}{24} \cdot \frac{16}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( p^4-6p^3+11p^2-6p \right) \cdot 16 }{ 24 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 16p^4-96p^3+176p^2-96p }{ 24 } \end{aligned} $$

Add $1-2p$ and $ \dfrac{4p^2-4p}{2} $ to get $ \dfrac{ \color{purple}{ 4p^2-8p+2 } }{ 2 }$.

Step 1: Write $ 1-2p $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 2 }$.

$$ \begin{aligned} 1-2p+ \frac{4p^2-4p}{2} & \xlongequal{\text{Step 1}} \frac{1-2p}{\color{red}{1}} + \frac{4p^2-4p}{2} = \frac{ \left( 1-2p \right) \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} + \frac{ 4p^2-4p }{ 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2-4p } }{ 2 } + \frac{ \color{purple}{ 4p^2-4p } }{ 2 }=\frac{ \color{purple}{ 4p^2-8p+2 } }{ 2 } \end{aligned} $$

Multiply $ \dfrac{p^3-3p^2+2p}{6} $ by $ -8 $ to get $ \dfrac{ -8p^3+24p^2-16p }{ 6 } $.

Step 1: Write $ -8 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{p^3-3p^2+2p}{6} \cdot -8 & \xlongequal{\text{Step 1}} \frac{p^3-3p^2+2p}{6} \cdot \frac{-8}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( p^3-3p^2+2p \right) \cdot \left( -8 \right) }{ 6 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -8p^3+24p^2-16p }{ 6 } \end{aligned} $$

Multiply $ \dfrac{p^4-6p^3+11p^2-6p}{24} $ by $ 16 $ to get $ \dfrac{ 16p^4-96p^3+176p^2-96p }{ 24 } $.

Step 1: Write $ 16 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{p^4-6p^3+11p^2-6p}{24} \cdot 16 & \xlongequal{\text{Step 1}} \frac{p^4-6p^3+11p^2-6p}{24} \cdot \frac{16}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( p^4-6p^3+11p^2-6p \right) \cdot 16 }{ 24 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 16p^4-96p^3+176p^2-96p }{ 24 } \end{aligned} $$

Add $ \dfrac{4p^2-8p+2}{2} $ and $ \dfrac{-8p^3+24p^2-16p}{6} $ to get $ \dfrac{ \color{purple}{ -8p^3+36p^2-40p+6 } }{ 6 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 3 }$.

$$ \begin{aligned} \frac{4p^2-8p+2}{2} + \frac{-8p^3+24p^2-16p}{6} & = \frac{ \left( 4p^2-8p+2 \right) \cdot \color{blue}{ 3 }}{ 2 \cdot \color{blue}{ 3 }} + \frac{ -8p^3+24p^2-16p }{ 6 } = \\[1ex] &=\frac{ \color{purple}{ 12p^2-24p+6 } }{ 6 } + \frac{ \color{purple}{ -8p^3+24p^2-16p } }{ 6 }=\frac{ \color{purple}{ -8p^3+36p^2-40p+6 } }{ 6 } \end{aligned} $$

Multiply $ \dfrac{p^4-6p^3+11p^2-6p}{24} $ by $ 16 $ to get $ \dfrac{ 16p^4-96p^3+176p^2-96p }{ 24 } $.

Step 1: Write $ 16 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{p^4-6p^3+11p^2-6p}{24} \cdot 16 & \xlongequal{\text{Step 1}} \frac{p^4-6p^3+11p^2-6p}{24} \cdot \frac{16}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( p^4-6p^3+11p^2-6p \right) \cdot 16 }{ 24 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 16p^4-96p^3+176p^2-96p }{ 24 } \end{aligned} $$

Add $ \dfrac{-8p^3+36p^2-40p+6}{6} $ and $ \dfrac{16p^4-96p^3+176p^2-96p}{24} $ to get $ \dfrac{ \color{purple}{ 16p^4-128p^3+320p^2-256p+24 } }{ 24 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 4 }$.

$$ \begin{aligned} \frac{-8p^3+36p^2-40p+6}{6} + \frac{16p^4-96p^3+176p^2-96p}{24} & = \frac{ \left( -8p^3+36p^2-40p+6 \right) \cdot \color{blue}{ 4 }}{ 6 \cdot \color{blue}{ 4 }} + \frac{ 16p^4-96p^3+176p^2-96p }{ 24 } = \\[1ex] &=\frac{ \color{purple}{ -32p^3+144p^2-160p+24 } }{ 24 } + \frac{ \color{purple}{ 16p^4-96p^3+176p^2-96p } }{ 24 } = \\[1ex] &=\frac{ \color{purple}{ 16p^4-128p^3+320p^2-256p+24 } }{ 24 } \end{aligned} $$